How much of the CO2 formed apparently dissolved in the water of the pneumatic trough?

Question

The carbon dioxide from the combustion of 1.50 grams of C2H6 is collected over water at 25 degrees Celsius. The pressure of CO2 in the collection flask is 746 mm Hg, and the colume is 2.00 L.

Answer 1

C2H6 + 3.5O2 ==> 2CO2 + 3H2O
So the combustion of one mole C2H6 would generate two moles CO2.
The molar mass of C2H6 is: 30.07 g/mol.
1.50g C2H6 = 0.0499 mol C2H6 ==> to generate 0.0998 mol CO2
At 25 degrees Celsius, the partial pressure of water vapor is 3.17kPa = 23.8 mm Hg. Thus the partial pressure of CO2 is about (746 - 23.8) Torr = 722 Torr. The number of moles of CO2 in the gas phase is:
n = PV/RT = (722/760)*2.00/(0.08206*298) = 0.0777 (mol)
The number of moles of CO2 dissolved in the water is:
0.0998mol - 0.0777 mol = 0.0221 mol.